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Please see attached picture to understand where I need help

Please see attached picture to understand where I need help-example-1

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We are given that two trains start a distance of 992 km apart. If they meet after 4 hours, this means that the sum of the distances of both trains after 4 hours traveling must be 992 km. This is written mathematically as:


d_a+d_b=992

Now, the distance traveled by an object is given by the following relationship:


\begin{gathered} d_a=r_at \\ d_b=r_bt \end{gathered}

Where:


\begin{gathered} r_a,r_b=\text{ rates of the trains} \\ t=\text{ time} \end{gathered}

Now, we substitute the values in the equation for the total distance:


r_at+r_bt=992

Now, the time is 4 hours, therefore, we have:


r_a(4)+r_b(4)=992

Now, we are also given that one train travels 20 km/h slower than the other. If a is the slower train then we have:


r_a=r_b-20

Substituting we get:


4(r_b-20)+4r_b=992

Now, we divide both sides by 4:


r_b-20+r_b=(992)/(4)

Solving the operations:


r_b-20+r_b=248

Now, we add like terms:


2r_b-20=248

Now, we add 20 to both sides:


\begin{gathered} 2r_b=248+20 \\ 2r_b=268 \end{gathered}

Now, we divide both sides by 2:


r_b=(268)/(2)=134

Therefore the fastest train has a rate of 134 km/h. Now, we determine the rate of the slower train:


\begin{gathered} r_a=134-20 \\ r_a=114 \end{gathered}

Therefore, the slower train travels ar 114 km/h

User Rob Van Der Veer
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